0.1 text{ mole} of potassium permanganate was | vedclass

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(A) The thermal decomposition of potassium permanganate is: $2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2 \uparrow$. Molar mass of $KMnO_4 = 39

Vedclass · Accepted Answer

$14.2$

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(A) The thermal decomposition of potassium permanganate is: $2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2 \uparrow$. Molar mass of $KMnO_4 = 39 + 55 + (16 	imes 4) = 158 	ext{ g/mol}$. Mass of $0.1 	ext{ mole}$ of $KMnO_4 = 0.1 	imes 158 = 15.8 	ext{ g}$. From the reaction stoichiometry,$2 	ext{ moles}$ of $KMnO_4$ produce $1 	ext{ mole}$ of $O_2$ gas. So,$0.1 	ext{ mole}$ of $KMnO_4$ produces $0.05 	ext{ mole}$ of $O_2$. Mass of $O_2$ gas $= 0.05 	imes 32 = 1.6 	ext{ g}$. Weight of residue = Initial weight - Weight of $O_2$ gas $= 15.8 - 1.6 = 14.2 	ext{ g}$.

$0.1 \text{ mole}$ of potassium permanganate was heated at $300^{\circ}C$. What is the weight (in $g$) of the residue? ($Mn = 55 \text{ u}$,$K = 39 \text{ u}$,$O = 16 \text{ u}$)

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